How do you integrate int ( x^2 + 7x +3) /( x^2 (x + 3))∫x2+7x+3x2(x+3) using partial fractions?

1 Answer
Narad T.
Jan 19, 2017

The answer is =-1/x+2ln(∣x∣)-ln(∣x+3∣)+C=−1x+2ln(∣x∣)−ln(∣x+3∣)+C

Explanation:

Let's perform the decomposition into partial fractions

(x^2+7x+3)/(x^2(x+3))=A/x^2+B/x+C/(x+3)x2+7x+3x2(x+3)=Ax2+Bx+Cx+3

=(A(x+3)+Bx(x+3)+Cx^2)/(x^2(x+3))=A(x+3)+Bx(x+3)+Cx2x2(x+3)

Equalising the denominators

x^2+7x+3=A(x+3)+Bx(x+3)+Cx^2x2+7x+3=A(x+3)+Bx(x+3)+Cx2

Let x=0x=0, =>⇒, 3=3A3=3A, =>⇒, A=1A=1

Let x=-3x=−3, =>⇒, -9=9C−9=9C, =>⇒, C=-1C=−1

Coefficients of x^2x2

1=B+C1=B+C, =>⇒, B=1-C=2B=1−C=2

Therefore,

(x^2+7x+3)/(x^2(x+3))=1/x^2+2/x-1/(x+3)x2+7x+3x2(x+3)=1x2+2x−1x+3

So,

int((x^2+7x+3)dx)/(x^2(x+3))=intdx/x^2+2intdx/x-intdx/(x+3)∫(x2+7x+3)dxx2(x+3)=∫dxx2+2∫dxx−∫dxx+3

=-1/x+2ln(∣x∣)-ln(∣x+3∣)+C=−1x+2ln(∣x∣)−ln(∣x+3∣)+C

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